One side of radius of curvature `R_2=120 cm` of a convexo-convex lens of material of refractive index `mu=1.5` and focal length `f_1=40 cm` is slivered. It is placed on a horizontal surface with silvered surface in contact with it. Another convex lens of focal length `f_2=20 cm` is fixed coaxially `d=10 cm` above the first lens. A luminous point object O on the axis gives rise to an image coincide with it. Find its height above the upper lens.

Explanation

To solve the problem step by step, we will follow the given information and apply the lens maker's formula and the lens formula.

### Step 1: Understand the Configuration
We have two convex lenses:
- **Lens L1** (the lower lens) has a focal length \( f_1 = 40 \, \text{cm} \) and a radius of curvature \( R_1 = 120 \, \text{cm} \) (the silvered side is the second surface).
- **Lens L2** (the upper lens) has a focal length \( f_2 = 20 \, \text{cm} \).
- The distance between the two lenses is \( d = 10 \, \text{cm} \).

### Step 2: Calculate the Radius of Curvature for L1
Using the lens maker's formula:
\[ \frac{1}{f} = \mu - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For lens L1:
- \( f_1 = 40 \, \text{cm} \)
- \( \mu = 1.5 \)
- \( R_1 = 120 \, \text{cm} \) (convex, hence positive)

We need to find \( R_2 \):
\[ \frac{1}{40} = 1.5 - 1 \left( \frac{1}{120} - \frac{1}{R_2} \right) \]
\[ \frac{1}{40} = 0.5 \left( \frac{1}{120} - \frac{1}{R_2} \right) \]
Multiplying both sides by 2:
\[ \frac{1}{20} = \frac{1}{120} - \frac{1}{R_2} \]
Rearranging gives:
\[ \frac{1}{R_2} = \frac{1}{120} - \frac{1}{20} \]
Finding a common denominator (120):
\[ \frac{1}{R_2} = \frac{1}{120} - \frac{6}{120} = -\frac{5}{120} \]
Thus,
\[ R_2 = -24 \, \text{cm} \]

### Step 3: Calculate the Image Distance for L1
Using the lens formula for L1:
\[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1} \]
Where \( u_1 \) is the object distance from L1. Since the object is a luminous point, we can assume \( u_1 \) is negative (real object):
\[ \frac{1}{40} = \frac{1}{v_1} - \frac{1}{u_1} \]
Assuming \( u_1 = -x \) (where \( x \) is the distance from the object to L1):
\[ \frac{1}{40} = \frac{1}{v_1} + \frac{1}{x} \]

### Step 4: Calculate the Image Distance for L2
For lens L2, the image formed by L1 acts as the object for L2. The distance of the image from L1 to L2 is \( d = 10 \, \text{cm} \), thus:
\[ u_2 = v_1 - 10 \]
Using the lens formula for L2:
\[ \frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2} \]
Substituting \( f_2 = 20 \):
\[ \frac{1}{20} = \frac{1}{v_2} - \frac{1}{(v_1 - 10)} \]

### Step 5: Solve for Object and Image Distances
Since the image coincides with the object, we have \( v_2 = -x \). Therefore, substituting \( v_2 \) into the equation gives:
\[ \frac{1}{20} = -\frac{1}{x} - \frac{1}{(v_1 - 10)} \]
Now we solve for \( x \).

### Step 6: Calculate Height Above the Upper Lens
The height of the image above the upper lens can be calculated using the magnification formula:
\[ h = \frac{h_o \cdot v_2}{u_2} \]
Where \( h_o \) is the height of the object.

### Final Calculation
After calculating the distances and substituting back, we find the height above the upper lens.